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4 years ago

(m•n) • p=m •(n•p)

Mathematics

2 answers:

Katena32 [7]4 years ago

5 0

The answer is C. Associative property of multiplication.

rosijanka [135]4 years ago

3 0

I think that the answer is C.

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HELP ASAP!!!!! Given: Point B is on the perpendicular bisector of AC¯¯¯¯¯. BD¯¯¯¯¯ bisects AC¯¯¯¯¯ at point D. Prove: B is equid

Harlamova29_29 [7]

Answer: Missing parts are,

In first blank, $AD\cong DC$ ,

In second blank, SAS postulate

In third blank, CPCTC postulate

Step-by-step explanation:

Since, Here D is the mid point on the line segment AC.

And BD is a perpendicular to the line AC.

Therefore, In triangles ADB and CDB ( shown in figure)

AD\cong DC ( By the definition of mid point)

$\angle BDA\cong \angle BDC$ ( right angles )

$BD\cong BD$ ( reflexive)

Thus, By SAS ( side angle side )postulate,

$\triangle ADB\cong \triangle CDB$

So, by CPCTC( Corresponding parts of congruent triangles are congruent)

$AB\cong CB$

Now, By definition of congruent segment,

AB=CB

By definition of equidistant,

B is equally far from both A and C.

7 0

3 years ago

Read 2 more answers

Need help on this asap

Ganezh [65]

(-5)(10)= 50 degrees

Brainliest?

5 0

3 years ago

One side of a square has a value of 3x+2, find the perimeter of the square

alexira [117]

Answer:

P = 12x +8

Step-by-step explanation:

The perimeter of a square is given by

P = 4s where s is the side length

P = 4(3x+2)

Distribute

P = 12x +8

6 0

4 years ago

Read 2 more answers

PLEASE HELP ASAP

aleksklad [387]

Answer:

5 units because 5 times 5 is 25

5 0

3 years ago

The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r

mash [69]

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$

3 0

3 years ago